Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $K$ be a field extension of $F$ with $[K:F]=m$. If $\gcd(n,m)=1$, show that $f$ is irreducible over $K$.

I thought suppose $f$ reducible over $K$, so $$f(x) = \underbrace{p(x)}_{(\deg = r)}\underbrace{q(x)}_{(\deg = s)}$$ where $1<r,s<n$ and both divides $n$. I imagine that $r$ or $s$ divides $m$ too, but I don't know how to prove this.

Any hint? Or is there a better way?


1 Answer 1


If $f=pq$ where $p$ is irreducible in $K[x]$ with degree $r$ and $0<r<n$.select a root $\alpha$ of $p$ .then $[K[\alpha]:K][K:F]=[K[\alpha]:F[\alpha]][F[\alpha]:F]$,that is $rm=[K[\alpha]:F[\alpha]]n$.Hence $n\mid r$ since n and m are coprime. Contradiction

  • $\begingroup$ ${}$Nice! Thanks! $\endgroup$
    – Lucas
    Commented May 4, 2018 at 6:27
  • $\begingroup$ Why you assumed p is irreducible in $K[x]$? Do we need only p to be irreducible or both of them or no one of them? $\endgroup$
    – Intuition
    Commented Feb 2 at 17:53

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